Pendulum Frequency Homework: 2.30 Hz

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The discussion revolves around calculating the frequency of a simple pendulum oscillating at 2.30 Hz under different acceleration conditions. It clarifies that the frequency will not remain constant when the pendulum accelerates upward or downward, as the net acceleration changes. The formula for frequency must incorporate the new net acceleration instead of the standard gravitational acceleration. Participants confirm that the length of the pendulum remains unchanged during these calculations. The correct approach involves adjusting the gravitational acceleration based on the direction of the pendulum's acceleration.
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Homework Statement



A simple pendulum oscillates with frequency 2.30 Hz.
(a) What is its frequency if it accelerates at 0.500 g upward?
(b) What is its frequency if it accelerates at 0.500 g downward?

Homework Equations



f=1/[2(pi)(L/g)^.5]


The Attempt at a Solution



Won't frequency be constant? So both a and b are 2.30Hz?
 
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If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.
 
G01 said:
If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.

So how about finding L? Before I just use f=2.3Hz and g=9.81 and solved for L. Is that correct?
 
Yes, The value of L found for when the pendulum is not accelerating will be the same in parts A and B. Nothing suggests the length of the string changes.
 
G01 said:
If the pendulum is accelerating upward or downward the net acceleration the pendulum is experiencing will change. So, in this case, where g normally is in the formula, you must replace with the new net acceleration, call it a. So, what would the net acceleration be for each case. If you can find these values, then you can find the frequencies.

So for part a g=9.81-0.5*9.81
and b would be g=9.81+0.5*9.81?

Thanks
 
That looks correct.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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